Re: Problem with "Which"

*To*: mathgroup at smc.vnet.net*Subject*: [mg122213] Re: Problem with "Which"*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Fri, 21 Oct 2011 06:22:49 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <201110201144.HAA05724@smc.vnet.net>*Reply-to*: drmajorbob at yahoo.com

Wait, there's a subtlety I got wrong, as the following shows: Quit {Which[f2 > 0, "T", f2 < 0, "H", f2 = 0, "S"], x, f} {Which[f2 > 0, "T", f2 < 0, "H", f2 = 0, "S"], x, f} f2 f2 f2 has no value, despite the "condition" f2 = 0 being reached. If f2 were set to 0 either before or after substituting x -> 0, substituting x -> 2 would give the very same result, and that's not what happened. Hence, there's an evaluation order issue I was missing. It would no longer matter, however, if you used f2 == 0 as you must have intended. Bobby On Thu, 20 Oct 2011 15:32:47 -0500, DrMajorBob <btreat1 at austin.rr.com> wrote: > In your Which statement, when x has not yet been substituted, neither f2 > > 0 nor f2 < 0 is True, so "f2 = 0" is reached. > > (You meant f2 == 0.) > > The value of f2 = 0 is zero, not True, so all three conditions have > failed and Which remains unevaluated. > > The value of ew (which probably should have been a localized Module > variable but isn't) is > > ew > > {{x -> 0}, {x -> 2}} > > When x -> 0, all three conditions fail again, since f2 = 0 still > resolves to 0, not True. > > When x -> 2, the condition f2 > 0 is True, so Which[...] becomes "T", > and that's what you're getting. > > In the If statement, f2 is still 0, having been set that way in the > Which statement, so Which resolves to "T". > > The code did what you told it to do. > > Bobby > > On Thu, 20 Oct 2011 06:44:54 -0500, mbmb <sb at 9y.com> wrote: > >> Who can check my module? >> >> KD[f0_, a0_, b0_] := Module[{f = f0, a = a0, b = b0}, >> f2 = D[f, {x, 2}]; >> ew = Solve[D[f, x] == 0, x, Reals]; >> Print["Extremwerte: ", {Which[f2 > 0, "T", f2 < 0, "H", f2 = 0, "S" ], >> x, f} /. ew]; >> Print["Extremwerte: ", {If[f2 > 0, "T", "H" ], x, f} /. ew]; >> ] >> >> When I enter: KD[x^3 - 3 x^2, -1, 4] the output is >> >> Extremwerte: {{Which[-6+6 x<0,H,f2=0,S],0,0},{T,2,-4}} >> >> whereas the IF-line gives >> >> Extremwerte: {{H,0,0},{T,2,-4}} >> >> Why can't I use Which in this case. Why doesn't Mathematica evaluate f2 >> in all cases? >> > > -- DrMajorBob at yahoo.com

**References**:**Problem with "Which"***From:*mbmb <sb@9y.com>